Problem 50 - College Algebra Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{g} .\) Determine the domain for each function. $$f(x)=\sqrt{x-5}, g(x)=\sqrt{5-x}$$

Step-by-Step Solution

Answer

The sum, difference, product and quotient of the functions are \(f(x) + g(x) = \sqrt{x - 5} + \sqrt{5 - x}\), \(f(x) - g(x) = \sqrt{x - 5} - \sqrt{5 - x}\), \(f(x)*g(x) = \sqrt{25 - x^2}\), \(\frac{f(x)}{g(x)} = \sqrt{\frac{x - 5}{5 - x}}\) . Their domains are \(x = 5\), \(x = 5\), \(5 \geq x \geq -5\) and \(x \gt 5\) respectively.

Step 1: Find the sum of the functions

The sum of the two functions, denoted \(f+g\), is given by adding the expressions for \(f(x)\) and \(g(x)\). So, \(f(x) + g(x) = \sqrt{x - 5} + \sqrt{5 - x}\)

Step 2: Find the difference of the functions

The difference of the two functions, denoted \(f - g\), is given by subtracting the expression for \(g(x)\) from \(f(x)\). So, \(f(x) - g(x) = \sqrt{x - 5} - \sqrt{5 - x}\)

Step 3: Find the product of the functions

The product of the two functions, denoted by \(fg\), is given by multiplying the expressions for \(f(x)\) and \(g(x)\). So, \(f(x)*g(x) = \sqrt{x - 5} * \sqrt{5 - x}, which simplifies to f(x)*g(x) = \sqrt{ (x - 5) * (5 - x) }, then to f(x)*g(x) = \sqrt{25 - x^2}\)

Step 4: Find the quotient of the functions

The quotient of the two functions, denoted by \(\frac{f}{g}\), is given by dividing the expression for \(f(x)\) by \(g(x)\). So, \(\frac{f(x)}{g(x)} = \frac{\sqrt{x - 5}}{\sqrt{5 - x}}\), which simplifies to: \(\frac{f(x)}{g(x)} = \sqrt{\frac{x - 5}{5 - x}}\) . Here, note that \(x \neq 5\) since that would make \(g(x)\) and thus the denominator, zero.

Step 5: Determine the domain

The domain of the individual functions \(f\) and \(g\) as well as the sum, difference, and product of these functions are determined by the restrictions of the square root as well as the undefined points of the quotient. The domain of \(f(x)\) is: \(x \geq 5\), \(g(x)\) is: \(x \leq 5\) . The domain of \(f(x) + g(x)\) and \(f(x) - g(x)\) are: \(x = 5\) as \(f(x)\) and \(g(x)\) share this common value. The domain of \(f(x)g(x)\) is: \(5 \geq x \geq -5\) as it removes the restriction of the square root. The domain of \(\frac{f(x)}{g(x)}\) is \(x \gt 5\) because it cannot be equal to \(5\) (denominator cannot be \(0\)).

Key Concepts

Domain of Functions

Understanding the domain of a function is crucial when dealing with mathematical expressions. The domain refers to all potential input values (x-values) that will output a valid number for the function.

For example, if we consider the function

• \(f(x) = \sqrt{x-5}\): The expression inside the square root, \(x - 5\), must be greater than or equal to zero. This means \(x \geq 5\). Therefore, the domain of \(f(x)\) consists of all numbers greater than or equal to 5.
• Similarly, for \(g(x) = \sqrt{5-x}\), the expression \(5-x\) must also be non-negative. Therefore, \(x \leq 5\), and thus the domain consists of numbers less than or equal to 5.

It's vital to analyze these specific conditions for each function to ensure that the domains don't yield undefined or imaginary numbers.

Sum of Functions

Adding two functions involves summing their expressions. This is straightforward, yet requires careful consideration of their respective domains.

For instance, when adding the functions

• \(f(x) = \sqrt{x-5}\) and \(g(x) = \sqrt{5-x}\), the result is \(f(x) + g(x) = \sqrt{x-5} + \sqrt{5-x} \).
• Now, the challenge is to figure out the domain where both these expressions are valid simultaneously.

Since the square roots have contradicting conditions, the only value of \(x\) that satisfies both functions is 5. Therefore, the domain of the sum is \(x = 5\). This unique scenario is where the functions overlap within their specified ranges.

Difference of Functions

The process of subtracting one function from another involves similar steps to adding, yet we pay careful attention to sign changes.

Let's consider the difference function:

• The expression \(f(x) - g(x) = \sqrt{x-5} - \sqrt{5-x}\) follows directly from the definitions of \(f(x)\) and \(g(x)\).
• As with addition, we need to find the overlapping domain. Both functions are defined independently, yet \(x = 5\) is the singular point satisfying both.

Thus, the domain for the difference is \(x = 5\) as well. This situation demonstrates that despite opposing operations, the domain remains constrained to overlapping limits.

Product of Functions

The product of two functions involves multiplying their expressions, demanding special attention to the resultant domain.

Take the functions:

• Multiplying \(f(x) = \sqrt{x-5}\) by \(g(x) = \sqrt{5-x}\) results in \(f(x) \cdot g(x) = \sqrt{(x-5)(5-x)}\), which simplifies to \(\sqrt{25-x^2}\).
• The structure inside the square root, \(25-x^2\), introduces a different domain. Since we take the square root, \(25-x^2\) cannot be negative, thus limiting \(x\) to values between \(-5\) and 5.

So, the product function's domain is \([-5, 5]\), broadening the possibilities while still constrained by square root rules.

Quotient of Functions

Dividing functions requires special care, especially regarding where values are undefined due to division by zero.

Consider the quotient function:

• Expressed as \(\frac{f(x)}{g(x)} = \frac{\sqrt{x-5}}{\sqrt{5-x}}\).
• The challenge arises from when \(g(x)\), the denominator, becomes zero, specifically when \(x = 5\). This results in an undefined value, notably a division by zero.

Consequently, the domain of the quotient function is \(x > 5\). This ensures the denominator is always positive, avoiding undefined operations and maintaining valid results.